Antisymmetric Tensors

Let’s continue yesterday’s project by considering antisymmetric tensors today. Remember that we’re starting with a tensor representation of $\mathrm{GL}(V)$ on the tensor power $V^{\otimes n}$ , which also carries an action of the symmetric group $S_n$ by permuting the tensorands. The antisymmetrizer (for today) is an element of the group algebra $\mathbb{F}[S_n]$ , and thus defines an intertwiner from $V^{\otimes n}$ to itself. Its image is thus a subrepresentation $A^n(V)$ of $\mathrm{GL}(V)$ acting on $V^{\otimes n}$ .

Now let’s again say $V$ has finite dimension $d$ and pick a basis $\{e_i\}$ of $V$ . Then we again have a basis of $V^{\otimes n}$ given by $n$ -tuples of basis elements of $V$ , and the permutation $\pi$ again acts by

$\displaystyle\pi\left(e_{i_1}\otimes...\otimes e_{i_n}\right)=e_{i_{\pi(1)}}\otimes...\otimes e_{i_{\pi(n)}}$

So let’s see what the antisymmetrizer looks like.

$\displaystyle\begin{aligned}A\left(e_{i_1}\otimes...\otimes e_{i_n}\right)=\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\pi\left(e_{i_1}\otimes...\otimes e_{i_n}\right)\\=\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)e_{i_{\pi(1)}}\otimes...\otimes e_{i_{\pi(n)}}\end{aligned}$

It’s just like the symmetrizer, except we have a sign in each term given by the signum representation of the symmetric group.

Immediately this tells us something very interesting: if the same basis element of $V$ shows up twice in the basic tensor, then the antisymmetrization of that tensor is automatically zero! This is because there’s some swap $\tau$ exchanging the two slots where we find the two copies. We can divide the permutations in $S_n$ into two collections: those which can be written as a series of swaps starting with $\tau$ and those which can’t. And throwing $\tau$ on at the beginning of the list just exchanges these two groups for each other in a bijection. If you don’t see that immediately, consider how it’s really the exact same thing as how we solved the airplane seat problem almost two years ago, and Susan’s related Thanksgiving seating problem. Then each permutation in one collection is paired with one in the other collection, and the two in a pair differ by only one swap. This means that one of them will get sent to $+1$ in the signum representation, and one will get sent to $-1$ . Since the permuted tensors they give are the same, the two terms will exactly cancel each other out, and we’ll just end up adding together a bunch of zeroes. Neat!

Just like for the symmetric tensors, many different basic tensors will antisymmetrize to the same basic antisymmetric tensor. Again, we use the unique permutation which puts the tensorands in order, but now we use the fact above: if any basis vector of $V$ is repeated, then the antisymmetrization is automatically zero. Thus we throw out all these vectors and only consider those $n$ -tuples $e_{i_1}\otimes...\otimes e_{i_n}$ with $i_1<...<i_n$ . This makes it really easy to count the dimension of $A^n(\mathbb{F}^d)$ , because this is just the number of subsets of cardinality $n$ of a set of cardinality $d$ .

As an example, let’s antisymmetrize $e_2\otimes e_1\otimes e_4$ . We write out a sum of all the permutations, with signs as appropriate:

$\displaystyle\begin{aligned}\frac{1}{3!}(-e_1\otimes e_2\otimes e_4+e_1\otimes e_4\otimes e_2+e_2\otimes e_1\otimes e_4\\-e_2\otimes e_4\otimes e_1-e_4\otimes e_1\otimes e_2+e_4\otimes e_2\otimes e_1)\end{aligned}$

We have no repeated terms here because if we did they’d pair off with opposite signs and cancel out. Now you might notice that we didn’t start with the tensorands in order. If we put them in order, we have to use an odd permutation to get $e_1\otimes e_2\otimes e_4$ . Then we’ll get the same result, but with the opposite signs:

$\displaystyle\begin{aligned}\frac{1}{3!}(e_1\otimes e_2\otimes e_4-e_1\otimes e_4\otimes e_2-e_2\otimes e_1\otimes e_4+\\e_2\otimes e_4\otimes e_1+e_4\otimes e_1\otimes e_2-e_4\otimes e_2\otimes e_1)\end{aligned}$

It’s a different result, yes, but it serves just as well as a basic antisymmetric tensor.

December 23, 2008 - Posted by John Armstrong | Algebra, Linear Algebra, Representation Theory

8 Comments »

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